LARAVEL UNIT TEST-seeInDatabase的对面 [英] LARAVEL UNIT TEST - Opposite of seeInDatabase
问题描述
在Laravel 5.1中,有一种方法可以使用seeInDatabase($ table,$ fields)...将某些数据存储在数据库中.
In Laravel 5.1 there is a method that asset if some data are in database using the seeInDatabase($table,$fields)...
是否可以断言数据库中是否有某些数据?类似于dontSeeInDatabase ...类似于 dontSeeJson >
Is there a way to assert if the some data arent in database? Something like dontSeeInDatabase... Similiar to dontSeeJson
推荐答案
Laravel v5.6
断言名称已更改
->assertDatabaseMissing(string $table, array $data, string $connection = null)
相反的是
->assertDatabaseHas(string $table, array $data, string $connection = null)
以前的Laravel版本
有两种方法:
->notSeeInDatabase($table, array $data)
和
->missingFromDatabase($table, array $data)
一个只是另一个的别名.
One is just an alias for the other.
有关可用测试方法的完整列表,请查看位于vendor/laravel/framework/src/Illuminate/Foundation/Testing
For a full list of available testing methods take a look at the traits located at vendor/laravel/framework/src/Illuminate/Foundation/Testing
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